I am trying to decrypt a file via brute-force. I am only worried about the ASCII characters 32-127(All possible keyboard characters). The key needs to be a two character key. I am starting my for-loop like this:
for(int i = 32; i < 128; i++){
//Key is a string already declared further up.
key = Character.toString((char) i);
}
I have printed that and it gives me all keys on the keyboard. I am stuck on how to make the two character string,
I have a few ideas in mind but they do not work. I thought about trying to just do something similar to this:
if(key.length() > 2)
key = "";
else
key += key(i);
but I think that will only give me characters that are "back-to-back" to each other.
I need to test for like 32 with 33-127 and so on and so for. Any insight would be highly appreciated. Thank you.
The best way to brute force this would be a nested for loop
for(int i = 32; i < 128; i++){
//Key is a string already declared further up.
for(int j = 32; j < 128; j++){
key = Character.toString((char) i) + Character.toString((char) j);
}
}
This says for every character, pair it with every character, so you would get all combinations of 2 character codes. You can do the checking right after you set the key
Did you try double for loop, something like this:
String key;
for(int i = 32; i < 128; i++){
for(int j = 32; j < 128; j++){
//Key is a string already declared further up.
key = (char)i+""+(char)j;
System.out.println("Key: " + key + ", " + key.length());
}
}
Related
Given a String, I want to create a frequency distribution of characters in the String. That is, for each distinct character in the string, I want to count how many times it occurs.
Output is a String that consists of zero or more occurrences of the pattern xd, where x is a character from the source String, and d is the number of occurrences of x within the String. Each x in the output should occur once.
The challenge is to do this without using an array or Collection.
Examples:
Source: "aasdddr" Result: "a2s1d3r1"
Source: "aabacc" Result: "a3b1c2"
Source: "aasdddraabcdaa" Result: "a6s1d4r1b1c1"
I tried this way:
String str = "aasdddr", result = "";
int counter = 0;
for(int i = 0; i < str.length(); i++){
result += "" + str.charAt(i);
for(int j = 1; j < str.length(); j++){
if(str.charAt(i) == str.charAt(j)){
counter++;
}
}
result += counter;
}
System.out.println(result);
My output is a1a2s3d6d9d12r13
Finally, I found the solution. But I think any question has more than one solution.
First, we should declare an empty string to keep the result. We use a nested loop because the outer loop will keep a character fixed during each iteration of the inner loop. Also, we should declare a count variable inside the outer loop. Because in each match, it will be increased by one and after controlling each character in the inner loop, it will be zero for the next check. Finally, after the inner loop, we should put a condition to check whether we have that character inside the result string. If there isn't any character like that, then it will be added to the result string. After that, its frequency (count) will be added. Outside of the loop, we can print it.
public class FrequenciesOfChar {
public static void main(String[] args) {
String str = "aabcccd"; // be sure that you don't have any digit in your string
String result = ""; // this will hold new string
for (int i = 0; i < str.length(); i++) { // this will hold a character till be checked by inner loop
int count = 0; // put here so that it can be zero after each cycle for new character
for (int j = 0; j < str.length(); j++) { // this will change
if(str.charAt(i) == str.charAt(j)){ // this will check whether there is a same character
count++; // if there is a same character, count will increase
}
}
if( !(result.contains(""+str.charAt(i))) ){ // this checks if result doesn't contain the checked character
result += ""+str.charAt(i); // first if result doesn't contain the checked character, character will be added
result += count; // then the character's frequency will be added
}
}
System.out.println(result);
}
}
Run Result:
aabcccd - a2b1c3d1
First, counter needs to be reset inside the for loop. Each time you encounter a character in the source String, you want to restart the counter. Otherwise, as you have seen, the value of the counter is strictly increasing.
Now, think about what happens if a character occurs in more than one place in the source String, as in the "aasdddraabcdaa" example. A sequence of 1 or more a appears in 3 places. Because, at the time you get to the 2nd occurrence of a, a has been previously counted, you want to skip over it.
Because the source String cannot contain digits, the result String can be used to check if a particular character value has already been processed. So, after fixing the problem with counter, the code can be fixed by adding these two lines:
if (result.indexOf (source.charAt(i)) >= 0) {
continue; }
Here is the complete result:
package stackoverflowmisc;
public class StackOverflowMisc {
public static String freqDist(String source) {
String result = "";
int counter ;
for (int i = 0; i < source.length(); i++) {
if (result.indexOf (source.charAt(i)) >= 0) { continue; }
counter = 1;
result += source.charAt(i);
for (int j = 1; j < source.length(); j++) {
if (source.charAt(i) == source.charAt(j)) {
counter++;
}
}
result += counter;
}
return result;
}
public static void main(String[] args) {
String [] test = {"aasdddr", "aabacc", "aasdddraabcdaa"};
for (int i = 0; i < test.length; ++i) {
System.out.println (test[i] + " - " + freqDist (test[i]));
}
System.out.println ("End of Program");
}
}
Run results:
aasdddr - a2s2d4r2
aabacc - a3b2c3
aasdddraabcdaa - a6s2d5r2b2c2
End of Program
In one of the Q&A comments, you said the source string can contain only letters. How would the program work if it were allowed to contain digits? You can't use the result String, because the processing inserts digits there. Again, this is an easy fix: Add a 3rd String to record which values have already been found:
public static String freqDist2(String source) {
String result = "", found = "";
int counter ;
for (int i = 0; i < source.length(); i++) {
if (found.indexOf (source.charAt(i)) >= 0) { continue; }
counter = 1;
result += source.charAt(i);
found += source.charAt(i);
for (int j = 1; j < source.length(); j++) {
if (source.charAt(i) == source.charAt(j)) {
counter++;
}
}
result += counter;
}
return result;
}
Another possibility is to delete the corresponding characters from the source String as they are counted. If you are not allowed to modify the Source String, make a copy and use the copy.
Comment: I don't know if this is what your professor or whomever had in mind by placing the "No array" restriction, because a String is essentially built on a char array.
I have a string is half-size font and i want to convert it to full size. I was tried to use this code
final String x = "01589846";
String b = "";
System.out.print("01589846");
int y = 0;
final char[] list = x.toCharArray();
for (int i = 0; i < list.length; i++) {
y = Integer.parseInt(String.valueOf(list[i]));
final char unicode = (char) (y + 65296);
b += unicode;
}
System.out.println(b);
}
it actually working but it only working with number.
Anyone have another way for this ? please help me !!!!!
Java Strings are Unicode. They don't need converting. Java does not natively use ASCII.
You apparently wish to map one set of Unicode characters to another. The appropriate tool for that would be a Map, but you'll have to populate the Map with your desired conversion taken from the Unicode code charts.
There may be some algorithmic way to do this for particular subranges; you seem to have discovered a way that works for (western) digits. Note that the fullwidth digits occupy codepoints 0xFF10 to 0xFF19, so the conversion formula is digit - '0' + 0xff10. 0xFF10 is 65296 decimal, but the hex is clearer, since it's what is used in published code charts.
Actually, it looks to me that the same thing works for all characters in the range SPACE to '~', presumably by design. Thus
for (int i=0; i<list.length; i++)
list[i] += 0xff00 - ' ';
Here, I simply assume without checking that list will only contain characters in the range of SPACE to '~', i.e., the Unicode range that corresponds to graphic (printable) ASCII characters. Dealing with other characters, for example Katakana, is more involved.
final String x = "012345 abcdef ABCDEF";
System.out.println(x);
String b = "";
final char[] list = x.toCharArray();
for (int i = 0; i < list.length; i++) {
if(Character.isDigit(list[i])) {
b += (char)(list[i] - 48 + 0xFF10);
} else if(Character.isUpperCase(list[i])) {
b += (char)(list[i] - 65 + 0xFF21);
} else if(Character.isLowerCase(list[i])) {
b += (char)(list[i] - 97 + 0xFF41);
} else if(Character.isWhitespace(list[i])) {
b += list[i];
} else {
b += (char)(list[i] - 33 + 0xFF01);
}
}
System.out.println(b);
Output:
012345 abcdef ABCDEF
０１２３４５ ａｂｃｄｅｆ ＡＢＣＤＥＦ
This is the code I am working upon. I dont know where I am going wrong.
package mcdcpairwise;
import java.io.*;
import java.util.*;
public class Permutation
{
public static void main(String[] args)
{
String a="000";
String b="|&";
for (int i=0; i < a.length(); i++){
if (i % 2 != 0){
a = a.substring(0,i-1) + b.substring(0,i-1). + a.substring(i, a.length()) + b.substring(i, b.length());
System.out.println(a);
}
}
}
}
The error I am facing is:
Exception in thread "main" java.lang.StringIndexOutOfBoundsException:
String index out of range: -2 at
java.lang.String.substring(String.java:1967) at
mcdcpairwise.Permutation.main(Permutation.java:13)
The output should be :
0|0&0
It isn't clear from your question exactly what your "rules" are for processing this. However, your output seems to simply insert a character between each character of your source a string.
Instead of using a substring, create a separate StringBuilder to add individual characters to. The code below produces the output you are looking for:
String string = "000";
StringBuilder output = new StringBuilder();
for (int i = 0; i < string.length(); i++) {
// Get current character in the string
char c = string.charAt(i);
// Add the current character to the output
output.append(c);
// If more characters exist, add the pipe
if (i != string.length() - 1) {
output.append("|");
}
}
System.out.println(output.toString());
The right code should be a.substring(0,i).
You can use String.toCharArray to get a char[] from a String. That way we can iterate more easily both String using an index.
String a="000";
String b="|&";
char[] arrayA = a.toCharArray();
char[] arrayB = b.toCharArray();
Then, all we have to do is to merge two array (from Strings) taking one character from both. Adding two conditions (one per array) to prevent any ArrayIndexOutOfBOundsException, we can insure we will merge two arrays.
StringBuilder sb = new StringBuilder();
//Add a char from both array (until we reach on of the limit)
int i = 0;
while( i < arrayA.length && i < arrayB.length){
sb.append(arrayA[i]).append(arrayB[i]);
++i;
}
Then we just need to add the remaining characters using a for loop on both arrays. Only one of those loop will be triggered (or none) since at least one previous condition (i < arrayA.length && i < arrayB.length) is already false.
//Add the rest of `a` if any
for(int j = i; j < arrayA.length; ++j){
sb.append(arrayA[j]);
}
//Add the rest of `b` if any
for(int j = i; j < arrayB.length; ++j){
sb.append(arrayB[j]);
}
System.out.println(sb.toString());
0|0&0
Here’s a one line solution:
System.out.println((a + b).replaceAll("(?<=.)(?=.{" + (a.length() - 1) + "}(.))|.(?=.{0," + (b.length() - 1) + "}$)", "$1"));
This works with all combinations of non-blank starting strings.
See live demo.
I am currently working on an assignment where I have to decrypt a secret phrase, but run and print 100 different possibilities based on what the ASCII value is. I have gotten this far and believe that what I have should work, but when I run the program it doesn't print anything, just says that the build is successful. Below is my code:
public static void main(String[] args) {
String phrase = ":mmZ\\dxZmx]Zpgy";
int value;
for (int key = 1; key >= 100; key++){
for (int i = 0; i > 15; i++ ){
int ascii = (int) phrase.charAt(i);
if(ascii + key > 126){
value = ((ascii + key) - 127) + 32;
System.out.print((char)value);
}
else{
value = (ascii + key);
System.out.print((char)value);
}
Thanks, any bit of information helps with this.
for (int key = 1; key >= 100; key++)
The for loop initializes key to 1, and then checks if key is greater than or equal to 100, which it isn't, so the loop immediately terminates.
Similarly, for (int i = 0; i > 15; i++ ) is checking if i is greater than 15 when it has just been initialized to 0.
I want to get the frequency of all 128 signs (ASCII) with the simplest code possible. No imports.
I am writing in Java (Eclipse), starting off like this:
public class Text {
public static void main (String[] args) {
then I want to calculate the frequency of each sign with a loop (preferably for loop). I know how to do this for a specific sign, e.g. the sign 'a' which is 97:
int a = 0;
for (int i = 0; i < s.length(); i++) { // s is a String
if (s.charAt(i) == 'a') {
a += 1;
}
}
System.out.println("a: " + a);
I need to create a table of all the signs (e.g. int[] p = new int p[1,2,3] - only for a string (or char?)) assign each index its number and then let a loop write out all the sign frequencies.
All this should be done only with loops and commands: .length, charAt().
Simply:
final String s = "Hello World!";
final int frequencies[] = new int[128];
for (int i = 0; i < s.length(); i++) {
final int ascii = (int) s.charAt(i);
frequencies[ascii]++;
}
(in response to user2974951's "answer")
That's the String representation of the array. Try printing with a loop instead:
for(int i = 0; i < frequencies.length; i++) {
System.out.println(frequencies[i]);
}
You can also try System.out.println(Arrays.toString(frequencies)); but that might look a bit ugly given the large amount of ASCII characters you are considering.